Question

I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to try to catch it. You are standing on the ground. Apparently, you do not want to stand directly under me; in fact, you would like to stand as far to one side as you can so that if I accidentally release it, it won’t hit you on the head. If you can run at 20 feet per second and I am at a height of 100 feet, how far away can you stand and still catch the egg if you start running when I let go?

Answer 1

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = $\sqrt{(100*2)/32.15}$ = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.